編譯技術(shù)第5次上機內(nèi)容

1、編譯技術(shù)第5次上機內(nèi)容目的：充分理解語義分析的方法及相關(guān)語義計算的執(zhí)行時機。要求：1以S屬性的語法制導定義為基礎(chǔ)，將下表的語義規(guī)則嵌套在語法分析的過程中，即實現(xiàn)語法制導的翻譯過程。產(chǎn) 生 式 語 義 規(guī) 則 L E n print (E.val) E E1 + T E.val := E1 .val + T.val E T E.val := T.val T T1 * F T.val := T1.val * F.val T F T.val := F.val F (E) F.val := E.val F digit F.val := digit.lexval2以詞法分析和語法分析部分的上機結(jié)果為基礎(chǔ)

2、，添加語義分析部分。即以LR文法為基礎(chǔ)。當進行產(chǎn)生式歸約時執(zhí)行對應(yīng)的語義動作。3輸入：5+3+8*2輸出：244 若輸入有誤，如：3+2則應(yīng)提示：重新輸入！5 由于輸入串是具體的數(shù)值，因此應(yīng)調(diào)用相應(yīng)的詞法分析的功能。/ Expression.cpp : Defines the entry point for the console application./#include stdafx.h#include conio.h#include#include#includeusing namespace std;#define L 0#define E 1#define T 2#define E_

3、 3#define T_ 4#define F 6#define digit 7/ 數(shù)字#define add 8/ 左括號#define mul 9/ 右括號#define lb 10#define rb11int nStackPtr;int Stack100;/ 棧void guiyue(stack &state,stack &value,char nex);void Push(int n)nStackPtr +;StacknStackPtr = n;void Pop()nStackPtr-;void PrintStack()int i;for (i = nStackPtr; i = 0;

4、 i-) if (Stacki = E) printf(E );if (Stacki = E_ ) printf(E );if (Stacki = T ) printf(T );if (Stacki = T_ ) printf(T );if (Stacki = F) printf(F );if (Stacki = digit) printf(digit );if (Stacki = add) printf(+ );if (Stacki = mul) printf(* );if (Stacki = lb) printf( );if (Stacki = rb) printf() );printf(

5、n);/ 利用棧來分析表達式串，判定表達式串是否正確/ /int main(int argc, char* argv)char strInput100;/ 存放表達式串bool bResult;int nInputPtr;nStackPtr = -1;nInputPtr = 0;bResult = true;/ 輸入表達式串，存放在 strInput中printf(請輸入表達式串：);scanf(%s,strInput);Push(E);PrintStack();while (bResult & nStackPtr = 0)switch(StacknStackPtr)case E:if (st

6、rInputnInputPtr = 0 & strInputnInputPtr = 0 & strInputnInputPtr = 0 & strInputnInputPtr= 0 & strInputnInputPtr=9) Pop();nInputPtr+;else bResult = false;PrintStack();break;case add:if (strInputnInputPtr = +) Pop();nInputPtr+;else bResult = false;PrintStack();break;case mul:if (strInputnInputPtr = *)

7、Pop();nInputPtr+;else bResult = false;PrintStack();break;case lb:if (strInputnInputPtr = () Pop();nInputPtr+;else bResult = false;PrintStack();break;case rb:if (strInputnInputPtr = ) Pop();nInputPtr+;else bResult = false;PrintStack();break;default: bResult = false; break;if (bResult = false) printf(

8、表達式有問題了n);elseprintf(表達式?jīng)]問題!n);/以上是詞法、語法分析代碼/-/以下是語義分析代碼stack state;stack value;int i=0;int size=strlen(strInput);for(i=0;i=0&strInputi=9)state.push(digit);value.push(strInputi-0);if(i!=(size-1)guiyue(state,value,strInputi+1);elseguiyue(state,value,#);else if(strInputi=+)state.push(add);else if(strI

9、nputi=*)state.push(mul);else if(strInputi=()state.push(lb);else if(strInputi=)state.push(rb);if(i!=(size-1)guiyue(state,value,strInputi+1);elseguiyue(state,value,#);/printf(Hello World!n);return 0;void guiyue(stack &state,stack &value,char nex)int next;if(nex=0&nex=9)next=digit;else if(nex=#)next=0;

10、elseswitch(nex)case +:next=add;break;case *:next=mul;break;case (:next=lb;break;case ):next=rb;break;stack state1=state;stack value1=value;int top_s1,top_s2;int top_v1;int top=state.top();bool flag=true;while(flag)switch(state.top()case digit:state.pop();state.push(F);state1=state;break;case F:if(st

11、ate.size()=3)state1.pop();top_s1=state1.top();state1.pop();top_s2=state1.top();state1.pop();if(top_s1=mul&top_s2=T)state1.push(T);state=state1;value1.pop();top_v1=value1.top();value1.pop();value1.push(top_v1*value.top();value=value1;elsestate.pop();state.push(T);state1=state;elsestate.pop();state.pu

12、sh(T);state1=state;break;case T:if(next=mul)flag=false;else if(state.size()=3)state1.pop();top_s1=state1.top();state1.pop();top_s2=state1.top();state1.pop();if(top_s1=add&top_s2=E)state1.push(E);state=state1;value1.pop();top_v1=value1.top();value1.pop();value1.push(top_v1+value.top();value=value1;elsestate.pop();state.push(E);state1=state;elsestate.pop();state.push(E);state1=state;break;case E:if(next=add|next=rb)flag=false;elseif(state.size()=1&state.top()=E)state.pop();state.push(L);printf(